Propellantless Field Propulsion and application

[Sith]

What about reentry speeds?

[Cinder]

If you can levitate yourself to orbit, you shouldn't have to deal with re-entry speeds any higher than current normal speeds.  At least not inherently due to ME propulsion.

[aero]

If you can levitate yourself to orbit, you shouldn't have to deal with re-entry speeds any higher than current normal speeds.  At least not inherently due to ME propulsion.

Well, if you are coming back from somewhere, surly you have some velocity. I guess the trick would be to shed the velocity before you hit the atmosphere. Would that take more thrust than simply escaping, or would it only take the same level of thrust? I expect you might plan your return/re-entry so that it took only as much thrust as you had available commensurate with the velocity you could tolerate on reaching atmosphere.

[Cinder]

Yeah, but none of that's due to using Mach Effect propulsion, which is IIRC the premise Sith was arguing from/asking about.  It's kind of vaguely phrased so I could be wrong.

[mlorrey]

Yeah, but none of that's due to using Mach Effect propulsion, which is IIRC the premise Sith was arguing from/asking about.  It's kind of vaguely phrased so I could be wrong.

Right, any ME drive capable of reaching such high speeds is also capable of decelerating from them, given sufficient power. Chemical rockets don't do this because chemical combustion is so inefficient that it puts severe mass penalties on such ships so as to make them infeasible. With ME thrusters, there is no propellant expulsion so there are no mass penalties. You can decelerate outside the atmosphere to subsonic or low supersonic speeds before reentry. This will also reduce G loads on reentry to something akin to riding an express elevator or being on approach to landing on an airliner.

[Cinder]

I'm just a layman here, but the intuitive POV I had was this:
If you can levitate to orbit, what's specifically happening is you're more-than-countering the planet's gravitational pull.  If you can make it to orbit this way, then you ought to have (assuming indefinite energy supply for the MET eqpt) strong enough propulsion to keep you there.  Even more so considering that gravity diminishes with distance.

Even if you had only very marginally better T/W than 1, you'd basically have a buoyant mass.  One that'd only get "lighter" the higher its altitude.  Which is equivalent to permanent station keeping, correct?  If that's the case, (and still assuming ME power supply is no issue) you wouldn't even need to have orbital velocity to stay in orbit, and therefore no re-rentry speed would be imposed other than reduced buoyancy to "sink" back to ground.

Of course it's immediately more complicated if power supply for such a floating access to orbit is unfeasible, if you still must rely on aerodynamic lift.

[Star-Drive]

Folks:

The size and complexity of the onboard electrical power supply required to drive an orbital M-E vehicle depends on the actual Newtons per Watt efficiency that the M-E thrusters in question can produce.  If we are talking about 1.0 milli-Newton per Watt or 1.0 Newton per Watt thrusters will tell us whether you need a 100 MWe nuclear reactor to drive the vehicle's thrusters, or just a good set of 100 kWe Li-poly batteries.  The M-E theory holds out the promise that the Li-poly battery solution will ultimately work, but the first several generations of M-E thrusters may need the light weight fusion reactor to drive it until we sort out the engineering optimizations required for the battery powered approach.  And that optimization task will most likely take a number of decades and a lot of development funding to get it right... 

[Dr_Fierro]

Dr. Fierro says: F=dP/dt is wrong, but that F= Mo*dV/dt is correct. This seemed an extraordinary statement to me because they ought to be identical statements, so I was looking for an explanation.

I was thinking if you start from Mo*V (the instantaneous definition of momentum) rather than than P (momentum), and then you differentiate WRT time to get F, you have started with the assumption that Mo does not change. Of course that eliminates any change in Mo. You've pulled it out of the integral! It seemed circular to me.


So I was hoping Dr. Fierro would explain why it is wrong to start from P rather than Mo*V, but perhaps I am not asking clearly because he appears to actually be starting, prima facie, from the assumption that V is the proper place to start rather than P.

cuddihy:

By definition P=Mo*V. The central issue is about the four-force definition: dP/dt or Mo*dV/dt? The equality only holds when you assume that Mo does not change, in which case we are dealing with a CLOSED physical system.  If we allow Mo to change, the physical system becomes an OPEN one and the four-force will depend on the definition you use.

The problem with the first definition (based upon dP/dt) is that, as a mathematical fact, in the Newtonian limit (c --> infinity) it predicts for zero external force that the particle's acceleration is frame dependent when Mo is allowed to change, so that the Galilean invariance is lost.

[cuddihy]

Dr. Fierro says: F=dP/dt is wrong, but that F= Mo*dV/dt is correct. This seemed an extraordinary statement to me because they ought to be identical statements, so I was looking for an explanation.

I was thinking if you start from Mo*V (the instantaneous definition of momentum) rather than than P (momentum), and then you differentiate WRT time to get F, you have started with the assumption that Mo does not change. Of course that eliminates any change in Mo. You've pulled it out of the integral! It seemed circular to me.


So I was hoping Dr. Fierro would explain why it is wrong to start from P rather than Mo*V, but perhaps I am not asking clearly because he appears to actually be starting, prima facie, from the assumption that V is the proper place to start rather than P.

cuddihy:

By definition P=Mo*V. The central issue is about the four-force definition: dP/dt or Mo*dV/dt? The equality only holds when you assume that Mo does not change, in which case we are dealing with a CLOSED physical system.  If we allow Mo to change, the physical system becomes an OPEN one and the four-force will depend on the definition you use.

The problem with the first definition (based upon dP/dt) is that, as a mathematical fact, in the Newtonian limit (c --> infinity) it predicts for zero external force that the particle's acceleration is frame dependent when Mo is allowed to change, so that the Galilean invariance is lost.

thanks for the reply!

[Star-Drive]

From Dr. Fierro:

"cuddihy:

"By definition P=Mo*V. The central issue is about the four-force definition: dP/dt or Mo*dV/dt? The equality only holds when you assume that Mo does not change, in which case we are dealing with a CLOSED physical system.  If we allow Mo to change, the physical system becomes an OPEN one and the four-force will depend on the definition you use.

The problem with the first definition (based upon dP/dt) is that, as a mathematical fact, in the Newtonian limit (c --> infinity) it predicts for zero external force that the particle's acceleration is frame dependent when Mo is allowed to change, so that the Galilean invariance is lost."


And here is Dr. Woodward's reponse in return:

"Paul:
 
In the limit c -> infinity it makes no sense to talk about four momentum.  In that limit space and time, and momentum and energy, are independent concepts.  To, suggest that Galilean invariance of Newtonian mechanics cannot accommodate systems with changing rest masses is, frankly, silly.
 
As for a particle accelerating with zero external force, and that the acceleration is frame dependent, nonsense.  In Newtonian mechanics particles don't accelerate in the absence of external forces, even if their rest masses are changing.  vdm/dt is not a force in the usual sense of forces.  It is a momentum flux that balances a real ma force in a closed system (so that the total momentum is conserved).  This sounds a lot like the sort of nonsense the Whealton was pushing a decade ago.  Sad to say, this stuff is not taught nearly well enough in typical mechanics classes.
 
P.S.  Rindler's definition of the four force as the derivative with respect to proper time of the four momentum is THE correct definition of the four force.  There is only one correct definition, and that is it.

Best,
 
Jim"

[cuddihy]

Thanks, Paul and to Dr Woodward as well. Lots to ponder...

[Star-Drive]

Thanks, Paul and to Dr Woodward as well. Lots to ponder...

Cuddihy:

While you are pondering here is a couple of links to Wolfgang Rindler's book on Special Relativity, first edition that was first published in 1985.  His four-force definition is in Chapter-5.

http://catatankuliah.com/2010/08/14/wolfgang-rindler-introduction-to-special-relativity/

Rindler's  improved 2nd edition, published in 1991, can be found at Amazon.com but it will set you back ~$55.00.

http://www.amazon.com/Introduction-Special-Relativity-Wolfgang-Rindler/dp/0198539525/ref=sr_1_1?s=books&ie=UTF8&qid=1291998555&sr=1-1

Woodward tells me that the 2nd edition has a much clearer discussion of the four-force than the first edition.

[cuddihy]

Thanks, Paul and to Dr Woodward as well. Lots to ponder...

Cuddihy:

While you are pondering here is a couple of links to Wolfgang Rindler's book on Special Relativity, first edition that was first published in 1985.  His four-force definition is in Chapter-5.

http://catatankuliah.com/2010/08/14/wolfgang-rindler-introduction-to-special-relativity/

Rindler's  improved 2nd edition, published in 1991, can be found at Amazon.com but it will set you back ~$55.00.

http://www.amazon.com/Introduction-Special-Relativity-Wolfgang-Rindler/dp/0198539525/ref=sr_1_1?s=books&ie=UTF8&qid=1291998555&sr=1-1

Woodward tells me that the 2nd edition has a much clearer discussion of the four-force than the first edition.

Thank you Paul

-tom

[MickQ]

Guys.  I have not been following this thread too closely so I'm not up to speed with much but a few questions.

1.  Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?
2.  Would/could it be classed simply as Anti Gravity ?
3.  As my main interest is Mars, would ME be a solution to landing large/heavy objects ?

Cheers for the season to ALL.

Mick.

[cuddihy]

Guys.  I have not been following this thread too closely so I'm not up to speed with much but a few questions.

1.  Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?
2.  Would/could it be classed simply as Anti Gravity ?
3.  As my main interest is Mars, would ME be a solution to landing large/heavy objects ?

Cheers for the season to ALL.

Mick.

Mick, merry Christmas to you.
1. Yes, with sufficient thrust, power to weight, and light enough batteries. But then, we don't currently know if that's possible with the materials we have.
2. no. The Woodward effect does not significantly cancel or even shield gravity in an observable manner, only inertia.
3. Yes, again but depends on thrust to weight. Look, where this technology is now can be compared to where nuclear power was in the 30s: wholly theoretical

[aceshigh]

Guys.  I have not been following this thread too closely so I'm not up to speed with much but a few questions.

1.  Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?
2.  Would/could it be classed simply as Anti Gravity ?
3.  As my main interest is Mars, would ME be a solution to landing large/heavy objects ?

Cheers for the season to ALL.

Mick.

2. no. The Woodward effect does not significantly cancel or even shield gravity in an observable manner, only inertia.

hmmm... your answer is correct, but its does gives the impression you are implying it can shield inertia on the whole vehicle... to be more clear to MickQ, the Woodward Effect only messes with inertia on the particles that it uses to propel the craft.

[D_Dom]

Guys.  I have not been following this thread too closely so I'm not up to speed with much but a few questions.

1.  Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?
2.  Would/could it be classed simply as Anti Gravity ?
3.  As my main interest is Mars, would ME be a solution to landing large/heavy objects ?

Cheers for the season to ALL.

Mick.

2. no. The Woodward effect does not significantly cancel or even shield gravity in an observable manner, only inertia.

hmmm... your answer is correct, but its does gives the impression you are implying it can shield inertia on the whole vehicle... to be more clear to MickQ, the Woodward Effect only messes with inertia on the particles that it uses to propel the craft.

Thanks for that clarification aceshigh. Let me check my understanding by discussion of another experiment Woodward has talked about, the air table demonstration of X/Y motion. If a demonstration were to be developed that carried internal batteries for power it would have a considerable mass.
 The ME thrusters would affect only the inertia of the particles used for propulsion. My question is assuming the particle inertia of the thrusters is small when compared to the inertia of the experiment including batteries, control circuits etc how much force can be generated using existing materials?

[aceshigh]

I assume that how much force it can generate is still open to debate, and even them are trying to discover precisely how much force they can get out of the device/effect.

I think it all depends of how much mass differential they can create between the "pushed" and the "pulled" particles... and how fast they can "push" them.

[cuddihy]

I assume that how much force it can generate is still open to debate, and even them are trying to discover precisely how much force they can get out of the device/effect.

I think it all depends of how much mass differential they can create between the "pushed" and the "pulled" particles... and how fast they can "push" them.

Well, sort of.

The problem is not "from the physics, how much force can you get out of it?," the physics says the amount of force output would be proportional to things like input frequency and power.

But so what? It's the wrong question to ask, it's kind of like asking, ok, based on what we know the nuclear binding energy to be, how much power can you get from a nuclear fission reactor? And the answer to that question is, "well, how big is the reactor? What's the distribution of the fissionable fuel? Which isotopes are they? What's those isotopes asorption & fission cross sections"...and on and on. You could ask 200 questions trying to lock down some input parameters for the answer to be non-variable.

And this is the same way. The physics says, given an oscillating bulk acceleration of so much and an oscillating change in energy of so much and such frequency, within so much mass, you should get a mass variation of such at   such frequency . Add in a "push heavy /pull light" external forcing mechanism, and you will get unbalanced force of such and such .

So the better question to ask is, in terms of these physics, what is the art of the possible given existing materials?

Look back through this thread, Paul March does quite a bit of explaining of what parameters would need to be to set a thrust-to-weight of greater than 1. Although something seems to have changed recently with regard to how to actually construct a Mach Effect thruster. I'm kind of waiting to see what happens with that myself...

Merry Christmas everyone.

[KelvinZero]

So the better question to ask is, in terms of these physics, what is the art of the possible given existing materials?

But the best question is always "Does this effect exist?"

I often get the impression that people on this thread do not realize what a hot topic this effect would be among physicists, even if it were a million times too small for any practical application, if it could just be demonstrated to actually exist.